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A speck of dirt is embedded 3.50 cm below the surface of a sheet of ice $(n = 1.3092)$. What is its apparent depth when viewed at normal incidence?
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Physics 103
Chapter 34
Geometric Optics
Section 3
Refraction at a Spherical Surface
Reflection and Refraction of Light
Wave Optics
Simon Fraser University
Hope College
University of Sheffield
Lectures
02:51
In physics, wave optics is the study of the behavior of light, or other electromagnetic waves, as they interact with matter. It is a branch of classical optics. While the term "optics" usually refers to the study of light propagation in free space, "wave optics" is used when it is important to take into account the effects of the material the light is traveling through. Wave optics is mostly concerned with the behavior of waves near the boundary between different media. When waves from a single source interact with each other their behavior can be difficult to predict. For example, two waves of the same frequency and wavelength can interfere, leading to a phenomenon known as interference fringes.
02:30
In optics, ray optics is a geometric optics method that uses ray tracing to model the propagation of light through an optical system. As in all geometric optics methods, the ray optics model assumes that light travels in straight lines and that the index of refraction of the optical material remains constant throughout the system.
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Okay, so we're doing Chapter 34 problems. 17 year. And this says a speck of dirt is embedded 3.5 centimeters below the surface of a sheet of ice, which is in with an index of refraction of 1.309 And we won't figure out what is its apparent death when viewed at normal incidence. Cool. Okay, so this is a drawing. Here we have. So we have. This is line. Here's our interface between the air in the ice. And we noted the air and extra traction is just one ices gave us was 1.309 here. So this bottom dot here is what we're gonna represent as the speck of dirt. So this is our object here, and these distances to the ice or the interface es and s prime are often object and image distance. So how we want to first do this set this up is we need to know that this is a flat surface. So we know that art is infinity for the radius of coverage. Cool. So we also need to figure out what is the quantity we're trying to find out. We're trying to find out this magnitude, Esperon, Because this is what the apparent death iss. So this is the apparent death. Because this is the what the depth is looks like from the Observer. So So now we need to figure out how to relate these together. And we can use our Lindsay collusion for with a nexus of refraction. And we know that that's gonna be given as the first index of fracturing over the object distance plus the second over, the image Nicholas Prime actually in this equals the difference in Exeter. Fractions over are Okay, so we already know. What are you going to infinity? And we know we want to find the magnitude of s prime so we can simplify this for a little bit of art. Infinitely, that means the right side of the equation is just zero. So now we can simply rearrange for s prime when this is gonna be negative in B s. Over didn't, eh, Cool. So that's a simple as we have to do. So let's just plug in our equation now, so we should make it a one times 3.5 centimeters all over 1.309 and that should give us a s prime of negative 2.67 since it's a virtual image. Centimeters. So the apparent depth, then, is the magnitude of this, since we can't be negative depth, and that's just 2.67 centimeters. Perfect. So what this is saying is that when the light goes from ice to air, it has been away from the normal, and the virtual image becomes closer to the surface than the object Awesome.
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